Limit of $frac{1}{7}e^{-2x^2}(1-4x^2)$ as $xtoinfty$












4












$begingroup$


I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function







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  • 6




    $begingroup$
    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    $endgroup$
    – Doug M
    Jan 3 at 15:45












  • $begingroup$
    Is $*$ multiplication or convolution in your question?
    $endgroup$
    – Asaf Karagila♦
    Jan 4 at 7:50
















4












$begingroup$


I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    $endgroup$
    – Doug M
    Jan 3 at 15:45












  • $begingroup$
    Is $*$ multiplication or convolution in your question?
    $endgroup$
    – Asaf Karagila♦
    Jan 4 at 7:50














4












4








4





$begingroup$


I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function







share|cite|improve this question











$endgroup$




I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function




limits exponential-function






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 7:51









Asaf Karagila♦

303k32429760




303k32429760










asked Jan 3 at 15:25









B. CzostekB. Czostek

556




556








  • 6




    $begingroup$
    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    $endgroup$
    – Doug M
    Jan 3 at 15:45












  • $begingroup$
    Is $*$ multiplication or convolution in your question?
    $endgroup$
    – Asaf Karagila♦
    Jan 4 at 7:50














  • 6




    $begingroup$
    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    $endgroup$
    – Doug M
    Jan 3 at 15:45












  • $begingroup$
    Is $*$ multiplication or convolution in your question?
    $endgroup$
    – Asaf Karagila♦
    Jan 4 at 7:50








6




6




$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45






$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45














$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50




$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50










5 Answers
5






active

oldest

votes


















6












$begingroup$

Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
    $endgroup$
    – Andreas Rejbrand
    Jan 4 at 7:01



















9












$begingroup$

Hint: Rewrite the expression as



$$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



Now, notice the growth of the numerator and denominator. Which grows more quickly?






share|cite|improve this answer











$endgroup$





















    5












    $begingroup$

    Your derivative is correct.



    Then arrange it as:



    $$frac{1-4x^2}{7e^{2x^2}}$$



    Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      Note that for $xge0$,
      $$
      begin{align}
      e^x
      &=1+x+frac{x^2}2+dots\
      &gefrac{x^2}2tag1
      end{align}
      $$
      Thus,
      $$
      e^{-2x^2}lefrac1{2x^4}tag2
      $$
      Applying $(2)$ to the expression for the derivative gives
      $$
      begin{align}
      left|frac17e^{-2x^2}!!left(1-4x^2right)right|
      &lefrac{4x^2+1}7frac1{2x^4}\
      &=frac2{7x^2}+frac1{14x^4}tag3
      end{align}
      $$






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



        I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
        $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you I completely forgot the L'Hopitals Rule
          $endgroup$
          – B. Czostek
          Jan 3 at 16:04










        • $begingroup$
          @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
          $endgroup$
          – Ethan Bolker
          Jan 4 at 1:34











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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Using L'Hôpital's rule , we get
        begin{align}
        lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
        &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
        end{align}        .






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          $endgroup$
          – Andreas Rejbrand
          Jan 4 at 7:01
















        6












        $begingroup$

        Using L'Hôpital's rule , we get
        begin{align}
        lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
        &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
        end{align}        .






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          $endgroup$
          – Andreas Rejbrand
          Jan 4 at 7:01














        6












        6








        6





        $begingroup$

        Using L'Hôpital's rule , we get
        begin{align}
        lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
        &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
        end{align}        .






        share|cite|improve this answer











        $endgroup$



        Using L'Hôpital's rule , we get
        begin{align}
        lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
        &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
        end{align}        .







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 15:39

























        answered Jan 3 at 15:32









        Thomas ShelbyThomas Shelby

        2,325221




        2,325221








        • 2




          $begingroup$
          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          $endgroup$
          – Andreas Rejbrand
          Jan 4 at 7:01














        • 2




          $begingroup$
          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          $endgroup$
          – Andreas Rejbrand
          Jan 4 at 7:01








        2




        2




        $begingroup$
        L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
        $endgroup$
        – Andreas Rejbrand
        Jan 4 at 7:01




        $begingroup$
        L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
        $endgroup$
        – Andreas Rejbrand
        Jan 4 at 7:01











        9












        $begingroup$

        Hint: Rewrite the expression as



        $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



        Now, notice the growth of the numerator and denominator. Which grows more quickly?






        share|cite|improve this answer











        $endgroup$


















          9












          $begingroup$

          Hint: Rewrite the expression as



          $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



          Now, notice the growth of the numerator and denominator. Which grows more quickly?






          share|cite|improve this answer











          $endgroup$
















            9












            9








            9





            $begingroup$

            Hint: Rewrite the expression as



            $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



            Now, notice the growth of the numerator and denominator. Which grows more quickly?






            share|cite|improve this answer











            $endgroup$



            Hint: Rewrite the expression as



            $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



            Now, notice the growth of the numerator and denominator. Which grows more quickly?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 17:17

























            answered Jan 3 at 16:00









            KM101KM101

            5,9431523




            5,9431523























                5












                $begingroup$

                Your derivative is correct.



                Then arrange it as:



                $$frac{1-4x^2}{7e^{2x^2}}$$



                Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  Your derivative is correct.



                  Then arrange it as:



                  $$frac{1-4x^2}{7e^{2x^2}}$$



                  Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Your derivative is correct.



                    Then arrange it as:



                    $$frac{1-4x^2}{7e^{2x^2}}$$



                    Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






                    share|cite|improve this answer









                    $endgroup$



                    Your derivative is correct.



                    Then arrange it as:



                    $$frac{1-4x^2}{7e^{2x^2}}$$



                    Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 3 at 16:49









                    Rhys HughesRhys Hughes

                    5,4811528




                    5,4811528























                        4












                        $begingroup$

                        Note that for $xge0$,
                        $$
                        begin{align}
                        e^x
                        &=1+x+frac{x^2}2+dots\
                        &gefrac{x^2}2tag1
                        end{align}
                        $$
                        Thus,
                        $$
                        e^{-2x^2}lefrac1{2x^4}tag2
                        $$
                        Applying $(2)$ to the expression for the derivative gives
                        $$
                        begin{align}
                        left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                        &lefrac{4x^2+1}7frac1{2x^4}\
                        &=frac2{7x^2}+frac1{14x^4}tag3
                        end{align}
                        $$






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          Note that for $xge0$,
                          $$
                          begin{align}
                          e^x
                          &=1+x+frac{x^2}2+dots\
                          &gefrac{x^2}2tag1
                          end{align}
                          $$
                          Thus,
                          $$
                          e^{-2x^2}lefrac1{2x^4}tag2
                          $$
                          Applying $(2)$ to the expression for the derivative gives
                          $$
                          begin{align}
                          left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                          &lefrac{4x^2+1}7frac1{2x^4}\
                          &=frac2{7x^2}+frac1{14x^4}tag3
                          end{align}
                          $$






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            Note that for $xge0$,
                            $$
                            begin{align}
                            e^x
                            &=1+x+frac{x^2}2+dots\
                            &gefrac{x^2}2tag1
                            end{align}
                            $$
                            Thus,
                            $$
                            e^{-2x^2}lefrac1{2x^4}tag2
                            $$
                            Applying $(2)$ to the expression for the derivative gives
                            $$
                            begin{align}
                            left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                            &lefrac{4x^2+1}7frac1{2x^4}\
                            &=frac2{7x^2}+frac1{14x^4}tag3
                            end{align}
                            $$






                            share|cite|improve this answer









                            $endgroup$



                            Note that for $xge0$,
                            $$
                            begin{align}
                            e^x
                            &=1+x+frac{x^2}2+dots\
                            &gefrac{x^2}2tag1
                            end{align}
                            $$
                            Thus,
                            $$
                            e^{-2x^2}lefrac1{2x^4}tag2
                            $$
                            Applying $(2)$ to the expression for the derivative gives
                            $$
                            begin{align}
                            left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                            &lefrac{4x^2+1}7frac1{2x^4}\
                            &=frac2{7x^2}+frac1{14x^4}tag3
                            end{align}
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 3 at 20:10









                            robjohn♦robjohn

                            266k27304626




                            266k27304626























                                2












                                $begingroup$

                                You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                                I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                                $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Thank you I completely forgot the L'Hopitals Rule
                                  $endgroup$
                                  – B. Czostek
                                  Jan 3 at 16:04










                                • $begingroup$
                                  @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  $endgroup$
                                  – Ethan Bolker
                                  Jan 4 at 1:34
















                                2












                                $begingroup$

                                You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                                I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                                $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Thank you I completely forgot the L'Hopitals Rule
                                  $endgroup$
                                  – B. Czostek
                                  Jan 3 at 16:04










                                • $begingroup$
                                  @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  $endgroup$
                                  – Ethan Bolker
                                  Jan 4 at 1:34














                                2












                                2








                                2





                                $begingroup$

                                You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                                I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                                $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






                                share|cite|improve this answer









                                $endgroup$



                                You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                                I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                                $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 3 at 15:42









                                AndrewAndrew

                                346115




                                346115












                                • $begingroup$
                                  Thank you I completely forgot the L'Hopitals Rule
                                  $endgroup$
                                  – B. Czostek
                                  Jan 3 at 16:04










                                • $begingroup$
                                  @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  $endgroup$
                                  – Ethan Bolker
                                  Jan 4 at 1:34


















                                • $begingroup$
                                  Thank you I completely forgot the L'Hopitals Rule
                                  $endgroup$
                                  – B. Czostek
                                  Jan 3 at 16:04










                                • $begingroup$
                                  @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  $endgroup$
                                  – Ethan Bolker
                                  Jan 4 at 1:34
















                                $begingroup$
                                Thank you I completely forgot the L'Hopitals Rule
                                $endgroup$
                                – B. Czostek
                                Jan 3 at 16:04




                                $begingroup$
                                Thank you I completely forgot the L'Hopitals Rule
                                $endgroup$
                                – B. Czostek
                                Jan 3 at 16:04












                                $begingroup$
                                @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                $endgroup$
                                – Ethan Bolker
                                Jan 4 at 1:34




                                $begingroup$
                                @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                $endgroup$
                                – Ethan Bolker
                                Jan 4 at 1:34


















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